A block of mass m is placed on a smooth wedge | vedclass

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(D) Consider the block of mass $m$ in the frame of the wedge. Since the block does not slip,it is in equilibrium in the frame of the wedge.The forces

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$m g \sec 	heta$

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(D) Consider the block of mass $m$ in the frame of the wedge. Since the block does not slip,it is in equilibrium in the frame of the wedge.The forces acting on the block are:$1$. Gravitational force $m g$ acting vertically downwards.$2$. Normal reaction $R$ exerted by the wedge,acting perpendicular to the inclined surface.$3$. Pseudo force $m a$ acting horizontally in the direction opposite to the acceleration of the wedge.Resolving the forces along the inclined plane:$m g \sin 	heta = m a \cos 	heta$$\Rightarrow a = g 	an 	heta$ ... $(i)$Resolving the forces perpendicular to the inclined plane:$R = m g \cos 	heta + m a \sin 	heta$Substituting the value of $a$ from equation $(i)$:$R = m g \cos 	heta + m (g 	an 	heta) \sin 	heta$$R = m g \cos 	heta + m g \frac{\sin^2 	heta}{\cos 	heta}$$R = \frac{m g \cos^2 	heta + m g \sin^2 	heta}{\cos 	heta}$$R = \frac{m g (\cos^2 	heta + \sin^2 	heta)}{\cos 	heta}$$R = \frac{m g}{\cos 	heta} = m g \sec 	heta$

$A$ block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be

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